# Upper triangular matrices and Billiard Arrays

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2 Y. Yang / Linear Algebra and its Applications 493 (2016) in Mat d+1 (F). View T as the transition matrix from a basis {u i } d i=0 of V to a basis {v i} d i=0 of V. Using T we construct three flags {U i } d i=0, {U i }d i=0, {U i }d i=0 on V as follows. For 0 i d, U i = Fu 0 + Fu Fu i = Fv 0 + Fv Fv i ; U i = Fu d + Fu d Fu d i ; U i = Fv d + Fv d Fv d i. In our first main result, we find a necessary and sufficient condition (called very good) on T for {U i } d i=0, {U i }d i=0, {U i }d i=0 to be totally opposite in the sense of [15, Definition 12.1]. In [15, Theorem 12.7] it is shown how three totally opposite flags on V correspond to a Billiard Array on V. Assume that the three flags {U i } d i=0, {U i }d i=0, {U i }d i=0 are totally opposite, and let B denote the corresponding Billiard Array on V. By [15, Lemma 19.1] B is determined up to isomorphism by a certain triangular array of scalar parameters called the B-values. In our second main result, we compute these B-values in terms of the entries of T. Let T d (F) denote the set of very good upper triangular matrices in Mat d+1 (F). Define an equivalence relation on T d (F) as follows. For T, T T d (F), we declare T T whenever there exist invertible diagonal matrices H, K Mat d+1 (F) such that T = HTK. In our third main result, we display a bijection between the following two sets: (i) the equivalence classes for on T d (F); (ii) the isomorphism classes of Billiard Arrays on V. We give a commutative diagram that illustrates our second and third main results. At the end of this paper, we give an example. In this example, we display a very good upper triangular matrix with entries given by q-binomial coefficients. We show that for the corresponding Billiard Array B, all the B-values are equal to q 1. The paper is organized as follows. Section 2 contains some preliminaries. Section 3 contains necessary facts about decompositions and flags. Section 4 is devoted to the correspondence between very good upper triangular matrices and totally opposite flags. This section contains our first main result. Section 5 contains necessary facts about Billiard Arrays. In Sections 6 8 we obtain our second and third main results. In Section 9, we display an example to illustrate our theory. 2. Preliminaries Throughout the paper, we fix the following notation. Let R denote the field of real numbers. Recall the ring of integers Z = {0, ±1, ±2,...} and the set of natural numbers N = {0, 1, 2,...}. Fix d N. Let {x i } d i=0 denote a sequence. We call x i the i-component of

3 510 Y. Yang / Linear Algebra and its Applications 493 (2016) the sequence. By the inversion of the sequence {x i } d i=0 we mean the sequence {x d i} d i=0. Let F denote a field. Let V denote a vector space over F with dimension d + 1. Let Mat d+1 (F) denote the F-algebra consisting of the d +1 by d +1matrices that have all entries in F. We index the rows and columns by 0, 1,..., d. Let I denote the identity matrix in Mat d+1 (F). 3. Decompositions and flags In this section, we review some basic facts about decompositions and flags. consisting of one- Definition 3.1. By a decomposition of V we mean a sequence {V i } d i=0 dimensional subspaces of V such that V = d i=0 V i (direct sum). Remark 3.2. For a decomposition of V, its inversion is a decomposition of V. Example 3.3. Choose a basis {f i } d i=0 of V. For 0 i d, define V i = Ff i. Then {V i } d i=0 is a decomposition of V. is in- Definition 3.4. Referring to Example 3.3, we say that the decomposition {V i } d i=0 duced by the basis {f i } d i=0. Definition 3.5. By a flag on V, we mean a sequence {W i } d i=0 of subspaces of V such that W i has dimension i +1for 0 i d and W i 1 W i for 1 i d. Example 3.6. Let {V i } d i=0 denote a decomposition of V. For 0 i d, define W i = V 0 + V V i. Then {W i } d i=0 is a flag on V. Definition 3.7. Referring to Example 3.6, we say that the flag {W i } d i=0 is induced by the decomposition {V i } d i=0. Definition 3.8. Consider a basis of V. That basis induces a decomposition of V, which in turn induces a flag on V. We say that flag is induced by the given basis. Lemma 3.9. (See [15, Section 6].) Suppose that we are given two flags on V, denoted by {W i } d i=0 and {W i }d i=0. Then the following are equivalent: (i) W i W j =0for i + j<d(0 i, j d); (ii) there exists a decomposition {V i } d i=0 of V that induces {W i} d i=0 and whose inversion induces {W i }d i=0. Moreover, suppose (i), (ii) hold. Then V i = W i W d i for 0 i d. are called op- Definition Referring to Lemma 3.9, the flags {W i } d i=0 and {W i }d i=0 posite whenever (i), (ii) hold.

4 Y. Yang / Linear Algebra and its Applications 493 (2016) We mention a variation on Lemma 3.9. Lemma (See [15, Section 6].) Suppose that we are given two flags on V, denoted by {W i } d i=0 and {W i }d i=0. Then they are opposite if and only if W i W j =0for i +j = d 1 (0 i, j d 1). Definition Suppose that we are given three flags on V, denoted by {W i } d i=0, {W i }d i=0, {W i }d i=0. These flags are said to be totally opposite whenever W d r W d s =0for all r, s, t (0 r, s, t d) such that r + s + t > d. W d t Lemma (See [15, Theorem 12.3].) Suppose that we are given three flags on V, denoted by {W i } d i=0, {W i }d i=0, {W i }d i=0. Then the following are equivalent: (i) The flags {W i } d i=0, {W i }d i=0, {W i }d i=0 are totally opposite. (ii) For 0 n d, the sequences {W i } d n i=0, {W d n W n+i }d n i=0, {W d n W n+i }d n i=0 are mutually opposite flags on W d n. (iii) For 0 n d, the sequences {W i }d n i=0, {W d n W n+i }d n i=0, {W d n W n+i} d n i=0 are mutually opposite flags on W d n. (iv) For 0 n d, the sequences {W i }d n i=0, {W d n W n+i} d n i=0, {W d n W n+i }d n i=0 are mutually opposite flags on W d n. For more information about flags, we refer the reader to [11] and [4]. 4. Upper triangular matrices and flags In this section, we explore the relation between upper triangular matrices and flags. First, we introduce some notation. Definition 4.1. For a matrix A Mat d+1 (F), we define some submatrices of A as follows. For 0 i j d, let A[i, j] denote the submatrix {A kl } 0 k j i,i l j of A. Note that A[0, d] = A. Definition 4.2. For a matrix A Mat d+1 (F) and 0 j d, we call the submatrix A[0, j] the j-th leading principal submatrix of A. Definition 4.3. For a matrix A Mat d+1 (F), we call it good whenever the submatrix A[i, d] is invertible for 0 i d. Definition 4.4. For a matrix A Mat d+1 (F), we call it very good whenever the submatrix A[i, j] is invertible for 0 i j d. Lemma 4.5. A matrix in Mat d+1 (F) is very good if and only if each of its leading principal submatrices is good.

5 512 Y. Yang / Linear Algebra and its Applications 493 (2016) Proof. By Definitions Referring to Definition 4.1, we now consider the case in which A is upper triangular. Lemma 4.6. For an upper triangular matrix A Mat d+1 (F), the submatrix A[0, j] is upper triangular for 0 j d. Proof. By Definition 4.1. Lemma 4.7. For an invertible upper triangular matrix A Mat d+1 (F), the submatrix A[0, j] is upper triangular and invertible for 0 j d. Proof. By Definition 4.1. Consider an invertible upper triangular matrix T Mat d+1 (F). View T as the transition matrix from a basis {u i } d i=0 of V to a basis {v i} d i=0 of V. Thus for 0 j d, v j = d T ij u i. (4.1) i=0 For the moment, pick x V. Then there exist scalars {b i (x)} d i=0 in F such that x = d b i (x)u i. (4.2) i=0 Moreover, there exist scalars {c i (x)} d i=0 in F such that By (4.1) (4.3), x = d c i (x)v i. (4.3) i=0 Tc = b, (4.4) where c = (c 0 (x), c 1 (x),..., c d (x)) t and b = (b 0 (x), b 1 (x),..., b d (x)) t. We now use T to construct three flags on V. Lemma 4.8. With the above notation, the following two flags on V coincide: (i) the flag induced by {u i } d i=0 ; (ii) the flag induced by {v i } d i=0.

6 Y. Yang / Linear Algebra and its Applications 493 (2016) Proof. By (4.1) and since T is upper triangular. We now define three flags on V, denoted by {U i } d i=0, {U i }d i=0, {U i }d i=0. The flag {U i } d i=0 is induced by the basis {u i} d i=0 or {v i} d i=0. The flag {U i }d i=0 (resp. {U i }d i=0 ) is induced by the basis {u d i } d i=0 (resp. {v d i} d i=0 ). More explicitly, for 0 i d, U i = Fu 0 + Fu Fu i = Fv 0 + Fv Fv i ; (4.5) U i = Fu d + Fu d Fu d i ; (4.6) U i = Fv d + Fv d Fv d i. (4.7) By Lemma 3.9, the flag {U i } d i=0 is opposite to the flags {U i }d i=0 and {U i }d i=0. Our next goal is to give a necessary and sufficient condition for the flags {U i }d i=0 and {U i }d i=0 to be opposite. We will use the following lemma. Lemma 4.9. With the above notation, for 0 i d 1, U i U d 1 i =0if and only if det(t [i +1, d]) 0. Proof. Consider x V. We refer to the notation around (4.2) and (4.3). We make two observations about x. The first observation is that by (4.6), we have x U i if and only if b n (x) = 0 for 0 n d 1 i. The second observation is that by (4.7), we have x U d 1 i if and only if c n(x) = 0 for 0 n i. In this case, by (4.4), T [i +1,d](c i+1 (x),c i+2 (x),...,c d (x)) t =(b 0 (x),b 1 (x),...,b d 1 i (x)) t. (4.8) First assume that det(t [i +1, d]) 0. We will show that U i U d 1 i = 0. To do this, we assume x U i U d 1 i, and show that x = 0. By the first observation, b n(x) = 0for 0 n d 1 i. By the second observation, c n (x) = 0for 0 n i and (4.8) holds. By these comments, T [i +1,d](c i+1 (x),c i+2 (x),...,c d (x)) t =0. (4.9) By (4.9) and det(t [i +1, d]) 0, the vector (c i+1 (x), c i+2 (x),..., c d (x)) t = 0. In other words, c n (x) = 0 for i +1 n d. We have shown that c n (x) = 0for 0 n d. Hence x = 0. Therefore U i U d 1 i = 0. Next assume that det(t [i + 1, d]) = 0. We will show that U i U d 1 i 0. By the assumption and linear algebra, there exists a nonzero vector w =(w i+1,w i+2,...,w d ) F d i such that T [i +1, d]w t =0. Choose the vector x such that c n (x) = 0for 0 n i and c n (x) = w n for i + 1 n d. Observe that x 0and satisfies (4.9). By the second observation, x U d 1 i and (4.8) holds. By (4.8) and (4.9), b n(x) =0for 0 n

7 514 Y. Yang / Linear Algebra and its Applications 493 (2016) d 1 i. By the first observation, x U i. We have shown that 0 x U i U Therefore U i U d 1 i 0. d 1 i. Proposition The flags {U i }d i=0 in the sense of Definition 4.3. and {U i }d i=0 are opposite if and only if T is good Proof. Recall from Definition 4.1 that T [0, d] = T. Since T is invertible, we obtain det(t [0, d]) 0. Therefore, by Definition 4.3, T is good if and only if det(t [i +1, d]) 0 for 0 i d 1. By Lemma 4.9, this happens if and only if U i U d 1 i =0for 0 i d 1. The result follows in view of Lemma Corollary T is good if and only if T 1 is good. Proof. Going through the construction around (4.5) (4.7) using T, we obtain a sequence of three flags {U i } d i=0, {U i }d i=0, {U i }d i=0. Repeating the construction with T replaced by T 1, we obtain a sequence of three flags {U i } d i=0, {U i }d i=0, {U i }d i=0. The result follows in view of Proposition Next, we give a necessary and sufficient condition for the three flags {U i } d i=0, {U i }d i=0, to be totally opposite in the sense of Definition {U i }d i=0 Proposition The three flags {U i } d i=0, {U i }d i=0, {U i }d i=0 are totally opposite in the sense of Definition 3.12 if and only if T is very good in the sense of Definition 4.4. Proof. By parts (i), (ii) of Lemma 3.13 along with Definition 4.4 and Lemma 4.5, it suffices to show that for 0 n d, the sequences {U i } d n i=0, {U d n U n+i }d n i=0, {U d n U n+i }d n i=0 are mutually opposite flags on U d n if and only if det(t [i, d n]) 0for 0 i d n. Let n be given. By (4.5) (4.7) we find that for 0 i d n, U d n U n+i = Fu d n i + Fu d n i Fu d n ; (4.10) U d n U n+i = Fv d n i + Fv d n i Fv d n. (4.11) By (4.5), (4.10) and Lemma 3.9, the flag {U i } d n i=0 on U d n is opposite to the flag {U d n U n+i }d n i=0 on U d n. Similarly, by (4.5), (4.11) and Lemma 3.9, the flag {U i } d n i=0 on U d n is opposite to the flag {U d n U n+i }d n i=0 on U d n. By Lemma 4.7, the submatrix T [0, d n] is invertible and upper triangular. Therefore we can apply Proposition 4.10 to the two flags {U d n U n+i }d n i=0 and {U d n U n+i }d n i=0 on U d n. By this, the two flags {U d n U n+i }d n i=0 and {U d n U n+i }d n i=0 are opposite if and only if det(t [i, d n]) 0 for 0 i d n. We have shown that the sequences {U i } d n i=0, {U d n U n+i }d n i=0, {U d n U n+i }d n i=0 are mutually opposite flags on U d n if and only if det(t [i, d n]) 0 for 0 i d n. The result follows. Corollary T is very good if and only if T 1 is very good.

8 Y. Yang / Linear Algebra and its Applications 493 (2016) Proof. Similar to the proof of Corollary Definition Let T d (F) denote the set of very good upper triangular matrices in Mat d+1 (F). Note that each element of T d (F) is invertible. 5. Billiard Arrays In this section, we develop some results about Billiard Arrays that will be used later in the paper. We will refer to the following basis for the vector space R 3 : e 1 = (1, 0, 0), e 2 = (0, 1, 0), e 3 = (0, 0, 1). Define a subset Φ = {e i e j 1 i, j 3, i j} of R 3. The set Φ is often called the root system A 2. For notational convenience define Note that α = e 1 e 2, β = e 2 e 3, γ = e 3 e 1. (5.1) Φ={±α, ±β,±γ}, α+ β + γ =0. Definition 5.1. Let d denote the subset of R 3 consisting of the three-tuples of natural numbers whose sum is d. Thus d = {(r, s, t) r, s, t N,r+ s + t = d}. Remark 5.2. We can arrange the elements of d in a triangular array. For example, if d = 3, the array looks as follows after deleting all punctuation: An element in d is called a location Definition 5.3. For η {1, 2, 3}, the η-corner of d is the location in d that has η-coordinate d and all other coordinates 0. By a corner of d we mean the 1-corner or 2-corner or 3-corner. The corners in d are listed below. de 1 =(d, 0, 0), de 2 =(0,d,0), de 3 =(0, 0,d).

9 516 Y. Yang / Linear Algebra and its Applications 493 (2016) Definition 5.4. For η {1, 2, 3}, the η-boundary of d is the set of locations in d that have η-coordinate 0. The boundary of d is the union of its 1-boundary, 2-boundary and 3-boundary. By the interior of d we mean the set of locations in d that are not on the boundary. Example 5.5. Referring to the picture of 3 from Remark 5.2, the 2-boundary of 3 consists of the four locations in the bottom row. Definition 5.6. For η {1, 2, 3} we define a binary relation on d called η-collinearity. By definition, locations λ, λ in d are η-collinear whenever the η-coordinate of λ λ is 0. Note that η-collinearity is an equivalence relation. Each equivalence class will be called an η-line. By a line in d we mean a 1-line or 2-line or 3-line. Example 5.7. Referring to the picture of 3 from Remark 5.2, the horizontal rows are the 2-lines of 3. Definition 5.8. Locations λ, µ in d are called adjacent whenever λ µ Φ. Definition 5.9. By an edge in d we mean a set of two adjacent locations. Definition By a 3-clique in d we mean a set of three mutually adjacent locations in d. There are two kinds of 3-cliques: (black) and (white). Lemma (See [15, Lemma 4.31].) Assume d 1. We describe a bijection from d 1 to the set of black 3-cliques in d. The bijection sends each (r, s, t) d 1 to the black 3-clique in d consisting of the locations (r +1, s, t), (r, s +1, t), (r, s, t +1). Lemma (See [15, Lemma 4.32].) Assume d 2. We describe a bijection from d 2 to the set of white 3-cliques in d. The bijection sends each (r, s, t) d 2 to the white 3-clique in d consisting of the locations (r, s +1, t +1), (r +1, s, t +1), (r +1, s +1, t). Lemma (See [15, Lemma 4.33].) For d, each edge is contained in a unique black 3-clique and at most one white 3-clique. Let P 1 (V )denote the set of 1-dimensional subspaces of V. Definition (See [15, Definition 7.1].) By a Billiard Array on V we mean a function B : d P 1 (V ), λ B λ that satisfies the following conditions: (i) for each line L in d the sum λ L B λ is direct; (ii) for each black 3-clique C in d the sum λ C B λ is not direct. We say that B is over F. We call V the underlying vector space. We call d the diameter of B.

10 Y. Yang / Linear Algebra and its Applications 493 (2016) Lemma (See [15, Corollary 7.4].) Let B denote a Billiard Array on V. Let λ, µ, ν denote the locations in d that form a black 3-clique. Then each of B λ, B µ, B ν is contained in the sum of the other two. Definition Let V denote a vector space over F with dimension d +1. Let B (resp. B ) denote a Billiard Array on V (resp. V ). By an isomorphism of Billiard Arrays from B to B we mean an F-linear bijection σ : V V that sends B λ B λ for all λ d. The Billiard Arrays B and B are called isomorphic whenever there exists an isomorphism of Billiard Arrays from B to B. From now until the end of Lemma 5.21, let B denote a Billiard Array on V. Definition Pick η {1, 2, 3}. Following [15, Section 9] we now define a flag on V called the B-flag [η]. For 0 i d, the i-component of this flag is λ B λ, where the sum is over all λ d that have η-coordinate at least d i. Definition Pick distinct η, ξ {1, 2, 3}. Following [15, Section 10] we now define a decomposition of V called the B-decomposition [η, ξ]. For 0 i d the i-component of this decomposition is the subspace B λ, where the location λ is described in the table below: η ξ λ i 1 2 (d i, i, 0) 2 1 (i, d i, 0) 2 3 (0,d i, i) 3 2 (0,i,d i) 3 1 (i, 0,d i) 1 3 (d i, 0,i) Lemma (See [15, Lemma 10.6].) For distinct η, ξ {1, 2, 3} the B-decomposition [η, ξ] of V induces the B-flag [η] on V. Lemma (See [15, Theorem 12.4].) The B-flags [1], [2], [3] on V from Definition 5.17 are totally opposite in the sense of Definition Lemma (See [15, Corollary 11.2].) Pick a location λ = (r, s, t) in d. Then B λ is equal to the intersection of the following three sets: (i) component d r of the B-flag [1]; (ii) component d s of the B-flag [2]; (iii) component d t of the B-flag [3]. Theorem (See [15, Theorem 12.7].) Suppose that we are given three totally opposite flags on V, denoted by {W i } d i=0, {W i }d i=0, {W i }d i=0. For each location λ = (r, s, t) in

11 518 Y. Yang / Linear Algebra and its Applications 493 (2016) d, define B λ = W d r W d s W d t. Then the map B : d P 1 (V ), λ B λ is a Billiard Array on V. Definition (See [15, Definition 8.1].) By a Concrete Billiard Array on V we mean a function B : d V, λ B λ that satisfies the following conditions: (i) for each line L in d the vectors {B λ } λ L are linearly independent; (ii) for each black 3-clique C in d the vectors {B λ } λ C are linearly dependent. We say that B is over F. We call V the underlying vector space. We call d the diameter of B. Example Let B denote a Billiard Array on V. For λ d pick 0 B λ B λ. Then the function B : d V, λ B λ is a Concrete Billiard Array on V. Definition Let B denote a Billiard Array on V, and let B denote a concrete Billiard Array on V. We say that B, B correspond whenever B λ spans B λ for all λ d. Definition Let λ, µ denote locations in d that form an edge. By Lemma 5.13 there exists a unique location ν d such that λ, µ, ν form a black 3-clique. We call ν the completion of the edge. From now until the end of Definition 5.33, let B denote a Billiard Array on V. Definition Let λ, µ denote locations in d that form an edge. By a brace for this edge, we mean a set of nonzero vectors u B λ, v B µ such that u + v B ν. Here ν denotes the completion of the edge. Lemma (See [15, Lemma 13.12].) Let λ, µ denote locations in d that form an edge. Then each nonzero u B λ is contained in a unique brace for this edge. Definition (See [15, Definition 14.1].) Let λ, µ denote adjacent locations in d. We define an F-linear map B λ,µ : B λ B µ as follows. For each brace u B λ, v B µ, the map B λ,µ sends u v. The map B λ,µ is well defined by Lemma Observe that B λ,µ : B λ B µ is bijective. Definition (See [15, Definition 14.9].) Let λ, µ, ν denote locations in d that form a white 3-clique. Then the composition Bλ,µ Bµ,ν B λ Bµ Bν Bν,λ Bλ is a nonzero scalar multiple of the identity map on B λ. The scalar is called the clockwise B-value (resp. counterclockwise B-value) of the clique whenever λ, µ, ν runs clockwise (resp. counterclockwise) around the clique.

12 Y. Yang / Linear Algebra and its Applications 493 (2016) Definition For each white 3-clique in d, by its B-value we mean its clockwise B-value. Definition By a value function on d, we mean a function ψ : d F \{0}. Definition (See [15, Definition 14.13].) Assume d 2. We define a function B : d 2 F as follows. Pick (r, s, t) d 2. To describe the image of (r, s, t) under B, consider the corresponding white 3-clique in d from Lemma The B-value of this 3-clique is the image of (r, s, t) under B. Observe that B is a value function on d 2 in the sense of Definition We call B the value function for B. Definition Let BA d (F) denote the set of isomorphism classes of Billiard Arrays over F that have diameter d. Definition Let VF d (F) denote the set of value functions on d. Definition Assume d 2. We now define a map θ : BA d (F) VF d 2 (F). For B BA d (F), the image of B under θ is the value function B from Definition Lemma (See [15, Lemma 19.1].) Assume d 2. Then the map θ : BA d (F) VF d 2 (F) from Definition 5.36 is bijective. Definition Let B denote a Concrete Billiard Array on V. Let B denote the corresponding Billiard Array on V from Definition Let λ, µ denote adjacent locations in d. Recall the bijection B λ,µ : B λ B µ from Definition Recall that B λ is a basis for B λ and B µ is a basis for B µ. Define a scalar B λ,µ F such that B λ,µ sends B λ B λ,µ B µ. Note that B λ,µ 0. Lemma (See [15, Lemma 15.6].) Let B denote a Concrete Billiard Array on V. Let λ, µ, ν denote locations in d that form a black 3-clique. Then B λ + B λ,µ B µ + B λ,ν B ν =0. Lemma (See [15, Lemma 15.9].) Let B denote a Concrete Billiard Array on V. Let B denote the corresponding Billiard Array on V from Definition Let λ, µ, ν denote the locations in d that form a white 3-clique. Then the clockwise (resp. counterclockwise) B-value of the clique is equal to B λ,µ Bµ,ν Bν,λ whenever the sequence λ, µ, ν runs clockwise (resp. counterclockwise) around the clique. Next we consider the 2-boundary of d.

13 520 Y. Yang / Linear Algebra and its Applications 493 (2016) Proposition Given a Billiard Array B on V, let {V i } d i=0 denote the B-decomposition [1, 3] of V from Definition Then for λ = (r, s, t) d, B λ V t + V t V d r. (5.2) Proof. We do induction on s. First assume that s = 0. Then by Definition 5.18, B λ = V t. Next assume that s > 0. Consider the black 3-clique in d with the locations λ = (r, s, t), µ = (r, s 1, t +1), ν =(r +1, s 1, t). By induction, By Lemma 5.15, B µ V t+1 + V t V d r ; (5.3) B ν V t + V t V d r 1. (5.4) B λ B µ + B ν. (5.5) The equation (5.2) follows from (5.3) (5.5). Lemma Let B denote a Billiard Array on V. Let {u i } d i=0 (resp. {v i} d i=0 ) denote a basis of V that induces the B-decomposition [1, 2] (resp. B-decomposition [1, 3]). Then the transition matrices between {u i } d i=0 and {v i} d i=0 are upper triangular. Proof. By Proposition We next show that the transition matrices in Lemma 5.42 are very good in the sense of Definition 4.4. By Corollary 4.13, it suffices to show that the transition matrix from {u i } d i=0 to {v i} d i=0 is very good. Lemma Referring to Lemma 5.42, let T denote the transition matrix from {u i } d i=0 to {v i } d i=0. Then T is very good in the sense of Definition 4.4. Proof. Consider the corresponding three flags {U i } d i=0, {U i }d i=0, {U i }d i=0 on V from (4.5) (4.7). By Lemma 5.19, the B-flag [1] (resp. [2]) (resp. [3]) is the flag {U i } d i=0 (resp. {U i }d i=0 ) (resp. {U i }d i=0 ). By Lemma 5.20, the three flags {U i} d i=0, {U i }d i=0, {U i }d i=0 are totally opposite. By Proposition 4.12, T is very good. Recall the set T d (F) from Definition Corollary Referring to Lemma 5.43, T T d (F). Proof. By Lemmas 5.42, 5.43.

14 Y. Yang / Linear Algebra and its Applications 493 (2016) Definition For location τ = (r, s, t) d 1, consider the corresponding black 3-clique in d from Lemma 5.11, with locations λ = (r +1, s, t), µ = (r, s + 1, t), ν =(r, s, t + 1). Given a Concrete Billiard Array B on V, we say that B is τ-standard whenever B λ B µ FB ν. We call B standard whenever B is τ-standard for all τ d 1. Let B denote a Billiard Array on V. For 0 i d, pick 0 f i B κ where κ = (d i, 0, i) d. Observe that {f i } d i=0 is a basis of V that induces the B-decomposition [1, 3]. Lemma With the above notation, there exists a unique standard Concrete Billiard Array B on V such that (i) B corresponds to B in the sense of Definition 5.25; (ii) for 0 i d, B κ = f i where κ = (d i, 0, i) d. Proof. First we show that B exists. Let λ = (r, s, t) d. We construct B λ by induction on s. For s = 0define B λ = f t. By the construction 0 B λ B λ. Next assume that s > 0. Consider the black 3-clique in d with the locations λ = (r, s, t), µ = (r, s 1, t + 1), ν =(r +1, s 1, t). The vectors B µ and B ν have been determined by the induction. By Lemma 5.28, there exists a nonzero vector B λ B λ such that B λ B ν FB µ. We have constructed B λ such that 0 B λ B λ for all λ d. By Example 5.24 and Definition 5.25, B is a Concrete Billiard Array on V that corresponds to B. By the above construction and Definition 5.45, B is standard. We have shown that B exists. The uniqueness of B follows by (ii) and Definition Consider the standard Concrete Billiard Array B from Lemma For location λ = (r, s, t) d, by Proposition 5.41 the vector B λ is a linear combination of f t, f t+1,..., f d r. Let {a i (λ)} d r i=t denote the corresponding coefficients, so that B λ = a t (λ)f t + a t+1 (λ)f t a d r (λ)f d r. (5.6) Lemma With the above notation, a t (λ) = 1. Proof. We do induction on s. First assume that s = 0. Then by Lemma 5.46(ii), B λ = f t. Therefore a t (λ) = 1. Next assume that s > 0. Consider the black 3-clique in d with the locations λ = (r, s, t), µ = (r, s 1, t +1), ν =(r +1, s 1, t). By Proposition 5.41, By induction, B ν = a t (ν)f t + a t+1 (ν)f t a d r 1 (ν)f d r 1 ; (5.7) B µ = a t+1 (µ)f t+1 + a t+2 (µ)f t a d r (µ)f d r. (5.8) a t (ν) =1. (5.9)

15 522 Y. Yang / Linear Algebra and its Applications 493 (2016) Since B is standard, B λ B ν FB µ. (5.10) By (5.6) (5.10) we have a t (λ) = 1. For more information about Billiard Arrays, we refer the reader to [15]. 6. Upper triangular matrices and Billiard Arrays Recall the set T d (F) from Definition In this section, we consider a matrix T T d (F). Using T we construct a Billiard Array B. Then for each white 3-clique in d, we compute its B-value in terms of the entries of T. Definition 6.1. Recall the set BA d (F) from Definition We define a map b : T d (F) BA d (F) as follows. Let T T d (F). View T as the transition matrix from a basis {u i } d i=0 of V to a basis {v i } d i=0 of V as around (4.1). Consider the corresponding three flags {U i } d i=0, {U i }d i=0, {U i }d i=0 on V from (4.5) (4.7). These flags are totally opposite by Proposition 4.12, so they correspond to a Billiard Array on V by Theorem Since the bases {u i } d i=0 and {v i} d i=0 are not uniquely determined, this Billiard Array is only defined up to isomorphism of Billiard Arrays. The isomorphism class of this Billiard Array is the image of T under b. In this section, we fix T T d (F). By Definition 4.14, T is upper triangular and invertible. Fix the two bases {u i } d i=0, {v i} d i=0 of V as around (4.1) and the three flags {U i} d i=0, {U i }d i=0, {U i }d i=0 on V from (4.5) (4.7). Let B denote the corresponding Billiard Array on V from Theorem Observe that B b(t ). Lemma 6.2. With the above notation, B is the unique Billiard Array on V such that for 0 i d, (i) B µ = Fu i where µ = (d i, i, 0) d ; (ii) B ν = Fv i where ν =(d i, 0, i) d. Proof. First we show that B satisfies (i). By Theorem 5.22, B µ = U i U d i U d. The subspace U i is given by (4.5). By (4.6), U d i = Fu d + Fu d Fu i. By (4.7), U d = Fv d + Fv d Fv 0 = V. By the above comments, U i U d i U d = Fu i. Hence B µ = Fu i. Therefore B satisfies (i). Similarly, B satisfies (ii). Next we show that B is the unique Billiard Array on V that satisfies (i) and (ii). Suppose that B is a Billiard Array on V that satisfies (i) and (ii). By Lemma 5.19, the B -flag [1] (resp. [2]) (resp. [3]) is the flag {U i } d i=0 (resp. {U i }d i=0 ) (resp. {U i }d i=0 ). By Lemma 5.21 and Definition 6.1, B λ = B λ for all λ d. Therefore B = B. We conclude that B is the unique Billiard Array on V that satisfies (i) and (ii).

16 Y. Yang / Linear Algebra and its Applications 493 (2016) Lemma 6.3. Consider the Billiard Array B on V. There exists a unique standard Concrete Billiard Array B on V such that (i) B corresponds to B in the sense of Definition 5.25; (ii) for 0 i d, B ν = v i where ν =(d i, 0, i) d. Proof. By Lemma 5.46 and Lemma 6.2(ii). Consider the Concrete Billiard Array B on V from Lemma 6.3. For location λ = (r, s, t) in d, by Proposition 5.41 the vector B λ is a linear combination of the vectors u s, u s+1,..., u d r and also a linear combination of the vectors v t, v t+1,..., v d r. For notational convenience, abbreviate b i (λ) = b i (B λ )in(4.2) and c i (λ) = c i (B λ )in(4.3), so that B λ = b s (λ)u s + b s+1 (λ)u s b d r (λ)u d r ; (6.1) B λ = c t (λ)v t + c t+1 (λ)v t c d r (λ)v d r. (6.2) In the following result we compute the coefficients in (6.2) in terms of the entries of T. The coefficients in (6.1) can be similarly computed, but we don t need these coefficients. Recall the T [i, j] notation from Definition 4.1. Proposition 6.4. With the above notation, for location λ = (r, s, t) in d c t (λ) = 1. Moreover, if s > 0, we have u = (T [t +1,d r]) 1 v, (6.3) where u = (c t+1 (λ), c t+2 (λ),..., c d r (λ)) t and v =(T 0t, T 1t,..., T s 1,t ) t. Proof. By Lemma 5.47 and Lemma 6.3, c t (λ) = 1. For the rest of the proof, assume that s > 0. By (6.1), b i (λ) = 0for 0 i s 1. By (6.2), c i (λ) = 0for 0 i t 1and d r +1 i d. Evaluating (4.4) using the above comments, we obtain v + T [t +1,d r]u =0. (6.4) The matrix T is very good by Definition 4.14, so T [t +1, d r] is invertible. Solving (6.4) for u, we obtain (6.3). Recall the B-value concept from Definition Our next goal is to compute these values for the Billiard Array B. Definition 6.5. Pick λ = (r, s, t) Z 3. If λ d, then define T [λ] to be the submatrix T [t, d r] from Definition 4.1. If λ / d, then define T [λ] to be the empty set.

17 524 Y. Yang / Linear Algebra and its Applications 493 (2016) For the rest of this section assume d 2. For a location τ =(r, s, t) d 2, consider the corresponding white 3-clique in d from Lemma This white 3-clique consists of the locations λ =(r, s +1,t+1), µ =(r +1,s,t+ 1), ν =(r +1,s+1,t). Next, consider the vectors µ ± α, µ ± β, µ ± γ, where α, β, γ are from (5.1). Note that λ = µ α and ν = µ + β. Moreover, µ + α =(r +2,s 1,t+1), µ α =(r, s +1,t+1), µ + β =(r +1,s+1,t), µ β =(r +1,s 1,t+ 2), µ + γ =(r, s, t +2), µ γ =(r +2,s,t). The above vectors form a hexagon as follows: µ + β µ α µ γ µ µ+ γ µ + α µ β Theorem 6.6. With the above notation, the B-value of the white 3-clique in d corresponds to τ is that where we interpret det( ) = 1. Proof. Consider the following vectors: det(t [µ + α]) det(t [µ + β]) det(t [µ + γ]) det(t [µ α]) det(t [µ β]) det(t [µ γ]), (6.5) µ α + β µ + β µ α µ γ µ µ+ γ We have three black 3-cliques with the following locations: (i) µ, µ α, µ + γ; (ii) µ γ, µ + β, µ; (iii) µ + β, µ α + β, µ α. For notational convenience, let τ denote the B-value of the white 3-clique in d that corresponds to τ. By Lemma 5.40, τ = B µ α,µ Bµ,µ+β Bµ+β,µ α. (6.6)

18 Y. Yang / Linear Algebra and its Applications 493 (2016) We now show that B µ α,µ = 1. (6.7) We apply Lemma 5.39 to the black 3-clique (i) and obtain For convenience we rewrite (6.8) as By Proposition 5.41, By Proposition 6.4, By (6.10), (6.11) and (6.13), By (6.9), (6.12) and (6.14), B µ α + B µ α,µ B µ + B µ α,µ+γ B µ+γ =0. (6.8) B µ α B µ +( B µ α,µ + 1)B µ + B µ α,µ+γ B µ+γ =0. (6.9) B µ Fv t+1 + Fv t Fv d r 1, (6.10) B µ α Fv t+1 + Fv t Fv d r, (6.11) B µ+γ Fv t+2 + Fv t Fv d r. (6.12) c t+1 (µ) =1, c t+1 (µ α) =1. (6.13) B µ α B µ Fv t+2 + Fv t Fv d r. (6.14) ( B µ α,µ + 1)B µ Fv t+2 + Fv t Fv d r. (6.15) By (6.10), the equation on the left in (6.13), and (6.15), we obtain (6.7). Next, we show that B µ,µ+β = det(t [µ + α]) det(t [µ]) det(t [µ β]) det(t [µ γ]). (6.16) There are two cases. First assume that s 0. We apply Lemma 5.39 to the black 3-clique (ii) and obtain By Proposition 5.41, B µ + B µ,µ+β B µ+β + B µ,µ γ B µ γ =0. (6.17) B µ γ Fv t + Fv t Fv d r 2. (6.18)

19 526 Y. Yang / Linear Algebra and its Applications 493 (2016) By (6.17) and (6.18), By Proposition 5.41, B µ + B µ,µ+β B µ+β Fv t + Fv t Fv d r 2. (6.19) By Proposition 6.4 and Cramer s rule, By (6.19) (6.21), B µ,µ+β = B µ Fv t+1 + Fv t Fv d r 1 ; B µ+β Fv t + Fv t Fv d r 1. (6.20) c d r 1 (µ) = ( 1)s det(t [t +1,d r 2]) ; det(t [t +2,d r 1]) c d r 1 (µ + β) = ( 1)s+1 det(t [t, d r 2]). (6.21) det(t [t +1,d r 1]) det(t [t +1,d r 2]) det(t [t +1,d r 1]). (6.22) det(t [t +2,d r 1]) det(t [t, d r 2]) Evaluating (6.22) using Definition 6.5, we obtain (6.16). Next assume that s = 0. Using an argument similar to (6.17) (6.21), we obtain Since we interpret det( ) = 1, B µ,µ+β = det(t [t +1,t+1]). (6.23) det(t [t, t]) det(t [µ + α]) = 1, det(t [µ β]) = 1. (6.24) Evaluate (6.23) using Definition 6.5. Combining the result with (6.24), we obtain (6.16). We have shown (6.16). In a similar manner using the black 3-clique (iii), we obtain B µ+β,µ α = det(t [µ + β]) det(t [µ + γ]). (6.25) det(t [µ α]) det(t [µ]) Evaluating (6.6) using (6.7), (6.16), (6.25) we obtain (6.5). 7. How T d (F) is related to BA d (F) Recall the set T d (F) from Definition 4.14, and the set BA d (F) from Definition In this section we explain how T d (F) is related to BA d (F). For convenience, we first consider an equivalence relation on T d (F).

20 Y. Yang / Linear Algebra and its Applications 493 (2016) Definition 7.1. For T, T T d (F), we declare T T whenever there exist invertible diagonal matrices H, K Mat d+1 (F) such that T = HTK. (7.1) The relation is an equivalence relation. For T T d (F), let [T ]denote the equivalence class of that contains T. Let T d (F) denote the set of equivalence classes for. Recall the map b : T d (F) BA d (F) from Definition 6.1. As we will see, b is surjective but not bijective. We will show that for T, T T d (F), b(t ) = b(t )if and only if T T. This tells us that the b-induced map T d (F) BA d (F) is bijective. Referring to Definition 7.1, assume that T T. Pick invertible diagonal matrices H, K Mat d+1 (F) that satisfy (7.1). Observe that the entries H ii 0and K ii 0for 0 i d. View T as the transition matrix from a basis {u i } d i=0 of V to a basis {v i} d i=0 of V as around (4.1). Recall the flags {U i } d i=0, {U i }d i=0, {U i }d i=0 from Definition 6.1. The matrix T can be viewed as the transition matrix from a basis {u i }d i=0 of V to a basis {v i }d i=0 of V, where u i = H ii u i and K iiv i = v i for 0 i d. Observe that the flags {U i } d i=0, {U i }d i=0, {U i }d i=0 are the same for T and T. Lemma 7.2. Let matrices T, T T d (F) satisfy T T in the sense of Definition 7.1. Then b(t ) = b(t ). Proof. By the discussion above the lemma statement, along with Definition 6.1. Definition 7.3. Using the map b : T d (F) BA d (F), we define a map b : T d (F) BA d (F) as follows. Given an equivalence class [T ] T d (F), the image of [T ]under b is b(t ). By Lemma 7.2 the map b is well-defined. Theorem 7.4. The map b : T d (F) BA d (F) from Definition 7.3 is bijective. Proof. First we show that the map b is surjective. Pick a Billiard Array B on V. It suffices to show that there exists T T d (F) such that B b(t ). For 0 i d, pick 0 u i B µ where µ = (d i, i, 0) d, and 0 v i B ν where ν =(d i, 0, i) d. By Definition 5.14, {u i } d i=0 and {v i} d i=0 are bases of V. Let T Mat d+1(f) denote the transition matrix from the basis {u i } d i=0 to the basis {v i} d i=0. By Corollary 5.44 T T d (F). Above Lemma 6.2 we refer to a Billiard Array obtained from Theorem By Lemma 6.2, this Billiard Array is the Billiard Array B. Hence B b(t )by the last sentence above Lemma 6.2. We have shown that the map b is surjective. Next we show that the map b is injective. Suppose that T, T T d (F) satisfy b(t ) = b(t ). We will show that T T in the sense of Definition 7.1. Define the Billiard Arrays B b(t )and B b(t )as around Definition 6.1. Since b(t ) = b(t ), the Billiard Arrays B and B are isomorphic. Therefore there exists an isomorphism σ of Billiard Arrays from B to B. By Definition 5.16, σ : V V is an F-linear bijection that sends

21 528 Y. Yang / Linear Algebra and its Applications 493 (2016) B λ B λ for all λ d. Associated with T we have the bases {u i } d i=0 and {v i} d i=0 of V from Definition 6.1. Similarly, associated with T we have bases {u i }d i=0 and {v i }d i=0 of V. To be more precise, T is the transition matrix from the basis {u i } d i=0 to the basis {v i } d i=0, and T is the transition matrix from the basis {u i }d i=0 to the basis {v i }d i=0. Since σ is an F-linear bijection, T is also the transition matrix from the basis {σ(u i )} d i=0 of V to the basis {σ(v i )} d i=0 of V. By Lemma 6.2, for 0 i d there exist nonzero h i, k i F such that σ(u i ) = h i u i and k iσ(v i ) = v i. Define diagonal matrices H, K Mat d+1(f) such that H ii = h i and K ii = k i for 0 i d. By construction H and K are invertible. By the above comments, T = HTK. By Definition 7.1, T T. We have shown that the map b is injective. Hence the map b is bijective. We continue to discuss the equivalence relation from Definition 7.1. Our next goal is to identify a representative in each equivalence class. Definition 7.5. A matrix T T d (F) is called nice whenever T 0i = T ii =1for 0 i d. Lemma 7.6. For the equivalence relation on T d (F) from Definition 7.1, each equivalence class contains a unique nice element in the sense of Definition 7.5. Proof. First, we show that each equivalence class in T d (F) contains at least one nice element. Choose T T d (F). By Definition 4.14, we have T 0i 0 and T ii 0 for 0 i d. Define diagonal matrices H, K Mat d+1 (F) such that H ii = T 0i /T ii and K ii =1/T 0i for 0 i d. By construction H and K are invertible. Let T = HTK. By construction, T T d (F) and T 0i = T ii =1for 0 i d. By Definition 7.1, we have T T. By Definition 7.5, T is nice. Therefore each equivalence class in T d (F) contains at least one nice element. Next, we show that this nice element is unique. Suppose that T, T are nice elements in T d (F) and T T. By Definition 7.1, there exist invertible diagonal matrices H, K Mat d+1 (F) that satisfy (7.1). For 0 i d, examining the (0, i)-entry in (7.1), we obtain T 0i = H 00T 0i K ii. Examining the (i, i)-entry in (7.1), we obtain T ii = H iit ii K ii. By Definition 7.5, we have T 0i = T ii = T 0i = T ii =1. By the above comments, we have K ii =1/H 00 and H ii = H 00 for 0 i d. Therefore H = H 00 I and K = I/H 00. Consequently T = T by (7.1). We have shown that each equivalence class in T d (F) contains a unique nice element. 8. A commutative diagram Recall the set T d (F) from Definition 4.14, the set BA d (F) from Definition 5.34, and the set VF d (F) from Definition For the moment assume d 2. In this section, we will describe how the sets T d (F), BA d (F), VF d (F) and VF d 2 (F) are related. In order to do this, we will establish a commutative diagram. As we proceed, some of our results do not require d 2. So until further notice, assume d 0. First we define a map D : T d (F) VF d (F).

22 Y. Yang / Linear Algebra and its Applications 493 (2016) Definition 8.1. For T T d (F), define a function D(T ) : d F as follows. For each location λ = (r, s, t) d, the image of λ under D(T )is det(t [λ]), where T [λ] is from Definition 6.5. Lemma 8.2. With reference to Definition 8.1, D(T ) is a value function on d in the sense of Definition In other words, D(T ) VF d (F). Proof. Pick λ =(r, s, t) d. By construction D(T )(λ) =det(t [t, d r]), which is nonzero since T is very good. We have shown that D(T )(λ) 0for all λ d. Therefore D(T ) VF d (F). Definition 8.3. We define a map D : T d (F) VF d (F) as follows. For T T d (F), the image of T under D is the function D(T )from Definition 8.1. By Lemma 8.2 the map D is well defined. For later use, we recall an elementary fact from linear algebra. Pick T Mat d+1 (F). For 0 i, j d, let T (i,j) denote the determinant of the d d matrix that results from deleting the i-th row and the j-th column of T. Applying the Laplace expansion to the bottom row of T we obtain det(t )= d ( 1) d+j T dj T (d,j). (8.1) Lemma 8.4. The map D : T d (F) VF d (F) from Definition 8.3 is bijective. j=0 Proof. For f VF d (F), we show that there exists a unique T T d (F) such that D(T ) = f. Our strategy is as follows. We are going to show that T exists. As we proceed, our calculation will show that there is only one solution for T. For 0 i, j d, we solve for T ij by induction on i +j. First assume that i +j =0, so that i = j = 0. Then D(T ) = f forces T 00 = f((d, 0, 0)). Next assume that i +j >0. If i > j, then T ij =0since T is required to be upper triangular. If i = 0, then D(T ) = f forces T 0j = f((d j, 0, j)). If 0 < i j, then D(T ) = f forces det(t [j i, j]) = f((d j, i, j i)). Consider the matrix T [j i, j]. The bottom right entry is T ij. All the other entries have already been computed by induction. We now compute det(t [j i, j]) using the Laplace expansion to its bottom row. Applying (8.1) to T [j i, j], we obtain a formula for det(t [j i, j]). In this formula the coefficient of T ij is det(t [j i, j 1]), which is nonzero by assumption. Therefore there is a unique solution for T ij. We have shown that the map D is bijective. From now until the end of Theorem 8.6, assume that d 2. Next we define a map w : VF d (F) VF d 2 (F).

23 530 Y. Yang / Linear Algebra and its Applications 493 (2016) Definition 8.5. Assume d 2. We define a map w : VF d (F) VF d 2 (F)as follows. Given f VF d (F), we describe the image of f under w. For a location τ =(r, s, t) d 2, the value of w(f)(τ) is f(µ + α)f(µ + β)f(µ + γ) f(µ α)f(µ β)f(µ γ). (8.2) Here µ = (r +1, s, t +1) d and α, β, γ are from (5.1). We interpret f(λ) = 1for all λ Z 3 such that λ / d. Theorem 8.6. Assume d 2. Then the following diagram commutes: T d (F) D VF d (F) b w BA d (F) θ VF d 2(F) Here θ is from Definition 5.36, b is from Definition 6.1, D is from Definition 8.3, and w is from Definition 8.5. Proof. For T T d (F) chase T around the diagram using Theorem 6.6. Recall the equivalence relation on T d (F) from Definition 7.1. Definition 8.7. Via the bijection D : T d (F) VF d (F), the equivalence relation on T d (F) induces an equivalence relation on VF d (F), which we denote by. In other words, for T, T T d (F), T T if and only if D(T ) D(T ). For f VF d (F), let [f] denote the equivalence class of that contains f. Let VF d (F) denote the set of equivalence classes for. We now give an alternative description of. Lemma 8.8. For f, f VF d (F) the following are equivalent: (i) f f in the sense of Definition 8.7; (ii) there exist nonzero h i, k i F (0 i d) such that for λ = (r, s, t) d. f (λ) =f(λ) s (h i k t+i ) (8.3) i=0

24 Y. Yang / Linear Algebra and its Applications 493 (2016) Proof. By Lemma 8.4 there exist T, T T d (F) such that D(T ) = f and D(T ) = f. (i) (ii) By Definition 8.7 we have T T. So there exist invertible diagonal matrices H, K Mat d+1 (F) that satisfy (7.1). Define h i = H ii and k i = K ii for 0 i d. By construction 0 h i, k i F for 0 i d, and T ij = h ik j T ij for 0 i, j d. Pick λ = (r, s, t) d. By Definition 8.1 and the above comments, f (λ) = det(t [t, d r]) = h 0 h 1...h s k t k t+1...k s+t det(t [t, d r]) = h 0 h 1...h s k t k t+1...k s+t f(λ). Therefore (8.3) holds. (ii) (i) Define diagonal matrices H, K Mat d+1 (F)such that H ii = h i and K ii = k i for 0 i d. By construction H, K are invertible. We will show that T = HTK. In order to do this, we show that T ij = h ik j T ij for 0 i, j d. We proceed by induction on i + j. First assume that i + j = 0, so that i = j = 0. Applying (8.3) to λ = (d, 0, 0) d we obtain T 00 = h 0 k 0 T 00. Next assume that i + j>0. If i >j, then T ij = T ij =0. Therefore T ij = h ik j T ij. If i = 0, then by applying (8.3) to λ = (d j, 0, j) d we obtain T 0j = h 0k j T 0j. If 0 < i j, then by applying (8.3) to λ = (d j, i, j i) d we obtain det(t [j i, j]) = det(t [j i, j]) i (h l k j i+l ). (8.4) By (8.4) and induction we routinely obtain T ij = h ik j T ij. We have shown that T = HTK. By Definition 7.1, T T. By Definition 8.7, f f. Definition 8.9. An element f VF d (F) is called fine whenever f((i, d i, 0)) = f((i, 0, d i)) = 1for 0 i d. l=0 Lemma For T T d (F) the following are equivalent: (i) T is nice in the sense of Definition 7.5; (ii) D(T ) is fine in the sense of Definition 8.9. Proof. By Definition 8.1, Similarly, by Definition 8.1, D(T )((i, d i, 0)) = T 00 T 11...T d i,d i (0 i d). (8.5) D(T )((i, 0,d i)) = T 0,d i (0 i d). (8.6)

25 532 Y. Yang / Linear Algebra and its Applications 493 (2016) First assume that T is nice. By Definition 7.5 along with (8.5) and (8.6), D(T )((i, d i, 0)) = D(T )((i, 0, d i)) =1for 0 i d. Therefore D(T )is fine in view of Definition 8.9. Next assume that D(T )is fine. By Definition 8.9 along with (8.5) and (8.6), T 0i = T ii =1for 0 i d. Therefore T is nice in view of Definition 7.5. Corollary Under the equivalence relation from Definition 8.7, each equivalence class contains a unique fine element in VF d (F). Proof. By Lemma 7.6 and Lemma Definition We define a map D : T d (F) VF d (F) as follows. Given an equivalence class [T ] T d (F), the image of [T ]under D is [D(T )]. By Definition 8.7 the map D is well-defined. Lemma The map D : T d (F) VF d (F) from Definition 8.12 is bijective. Proof. By Lemma 8.4 and Definition 8.7. For the rest of this section assume d 2. Lemma Assume d 2. Recall the map w from Definition 8.5. For f, f VF d (F), suppose that f f in the sense of Definition 8.7. Then w(f) = w(f ). Proof. Apply (8.2) to f and evaluate the result using (8.3). Definition Assume d 2. We define a map w : VF d (F) VF d 2 (F) as follows. Given an equivalence class [f] VF d (F), the image of [f] under w is w(f). By Lemma 8.14 the map w is well-defined. Lemma Assume d 2. Then the map w : VF d (F) VF d 2 (F) from Definition 8.15 is bijective. Proof. We will show that for g VF d 2 (F), there exists a unique fine f VF d (F) such that w(f) = g. Our strategy is as follows. We are going to show that f exists. As we proceed, our calculation will show that there is only one solution for f. For λ = (r, s, t) d, we solve for f(λ) by induction on s r. First assume that s r = d, so that λ = (d, 0, 0). By Definition 8.9, f(λ) = 1. Next assume that s r> d. If s = 0 or t = 0, then by Definition 8.9, f(λ) = 1. If s 0and t 0, then by Definition 8.5, g(τ) = f(µ + α)f(µ + β)f(µ + γ) f(µ α)f(µ β)f(µ γ), (8.7)

26 Y. Yang / Linear Algebra and its Applications 493 (2016) where τ =(r, s 1, t 1) d 2, µ = (r +1, s 1, t) d and α, β, γ are from (5.1). In terms of r, s, t the equation (8.7) becomes g(r, s 1,t 1) = f(r +2,s 2,t)f(r +1,s,t 1)f(r, s 1,t+ 1) f(r, s, t)f(r +1,s 2,t+1)f(r +2,s 1,t 1). (8.8) In the right hand side of (8.8), all the factors except for f(λ) = f(r, s, t) have been determined by induction, and these factors are nonzero. Hence f(λ) = f(r, s, t) is uniquely determined. We have shown that for g VF d 2 (F), there exists a unique fine f VF d (F) such that w(f) = g. By this and Corollary 8.11, the map w is bijective. Theorem Assume d 2. Then the following diagram commutes: T d (F) D VF d (F) b w BA d (F) θ VF d 2(F) Here θ is from Definition 5.36, b is from Definition 7.3, D is from Definition 8.12, and w is from Definition Proof. Use Theorem 8.6. Remark In the commutative diagram from Theorem 8.17, each map is bijective. 9. An example In this section, we will give an example to illustrate Theorem 6.6. We first recall some notation. Fix 0 q F. For n N define n 1 [n] q = q i, [n]! q = i=0 n [i] q. i=1 We interpret [0] q =0and [0]! q = 1. For n, k Z we define [ ] n k as follows. For 0 k n, q [ ] n = k q [n]! q [k]! q[n k]!. (9.1) q If k <0or k >n, then for notational convenience define [ ] n k q =0.

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